Help in math Please!!

Our lesson in "Adding and Subtracting Rational Expressions," I have already searched in the internet on how to solve, but the steps to get it is too hard for me can anyone help please D: just try to simplify the explaination because mostly the websites give complicated explanations to solving D:

this is just an example:

x 3x
-------------- + ------------------- (x2) is (X squared)
x2-7x+10 x2-3x-10


Im not trying to get answers for my homework its just the Steps, i dont know what to do first factor or something please hel;p me D:

So the math problem is x/(x^2-7x 10) 3x/(x^2-3-10)
first factor it.

x^2-7x 10

since we know that and is positice and - and - is positive
(x- 2 )(x- 5 )

numbers that multiple together to make 10 is 1*10 and 2*5
2*5 = 10 and 2 5 is 7 we can place in either parenthesis

Same for x^2-3x-10
since last is 10 we use same info 2*5
2*5 is 10 and -5 and 2 is 3 , so
(x 2 )(x- 5)
it should now look like:
x/(x-2)(x-5) 3x/(x 2)(x-5)
LCD= (x-5)(x-2)(x 2)
since left half has (x-2) and (x-5) we multiply both numerator and denominator by (x 2) and get x^2 2x
Right half has (x 2) and (x-5) so multiply numerator and denominator by (x-2) and get 3x^2-6x

Add x^2 2x 3x^2-6x and we get 4x^2-4x simplified is
4x(x-1)/(x-2)(x-5)(x 2)

YA has no math interface - it is very frustrating.for everyone
x/(x^2-7x+10) + 3x/(x^2-3x-10) =
x/(x-2)(x-5) + 3x/(x-5)(x+2)
LCD = (x-5)(x+2)(x-2)
So putting everything over the LCD
[x(x+2) + 3x(x-2)]/(x-5)(x+2)(x-2)=
[x^2+2x+3x^2-6x]/(x-5)(x+2)(x-2)=
[4x^2-4x]/(x-5)(x+2)(x-2)=
4(x^2-1)/(x-5)(x+2)(x-2)=
4(x+1)(x-1)/(x-5)(x+2)(x-2)